Rindler coordinates in spherical form

Question

From the Minkowski metric, it is almost straightforward to change coordinates and pass to Rindler metric $$ ds^2 = e^{2a\xi}(-d\tau^2 + d\xi^2) + dy^2 + dz^2 $$
given the transformation law $$ t = \frac{1}{a}e^{a\xi}\sinh{a\tau} $$ $$ x = \frac{1}{a}e^{a\xi}\cosh{a\tau} $$ At any fixed $\xi$, let's say $\xi = 0$, these coordinates represent the frame of an uniformly accelerated observer, with acceleration pointing in the x-direction. The question is how can i find a "polar" or rather "spherical" version of the Rindler coordinates describing the frame of an observer accelerating in the radial direction of a $(t,r,\theta,\phi)$ coordinate system? Writing Minkowski in spherical coordinates is very simple, but it is not trivial to make the same change for Rindler, due to this radial acceleration vs x-acceleration question.

Answer 1

If I understand correctly, let $(t,r,\theta,\phi)$ be the spherical coordinates entered about an arbitrary origin of an inertial frame, you want a new coordinate system $(\tau,\rho,\theta,\phi)$ such that at fixed $\rho_0,\theta_0,\phi_0$, the worldline parametrised by $\tau$ is the one of an accelerating observer at proper rate $a$ in the radial direction with $\tau$ their proper time. For $\rho$, the natural parametrisation would be the point of closest approach, but to keep your approach, I'll take a normalised log of the distance of closest approach.

You just need to calculate position of these accelerating observers: $$ t = \frac{1}{a}e^{a\rho_0}\sinh(a\tau) \\ r = \frac{1}{a}e^{a\rho_0}\cosh(a\tau) \\ \theta = \theta_0 \\ \phi = \phi_0 \\ $$ which gives you your coordinate change. Plugging it into the metric: $$ ds^2=-dt^2+dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi\\ =-e^{2a\rho}d\tau^2+e^{2a\rho}d\rho^2+\left(\frac{1}{a}e^{a\rho}\cosh(a\tau)\right)^2d\theta^2+\left(\frac{1}{a}e^{a\rho}\cosh(a\tau)\right)^2\sin^2\theta d\phi^2 $$

Taking instead the distance of closest approach, which is physically more transparent and the analogy between Rindler coordinates and polar coordinates is made more apparent, your original formulas become: $$ \tilde \xi = \frac{1}{a}e^{a\xi}\\ t = \tilde \xi\sinh(a\tau) \\ x = \tilde \xi\cosh(a\tau) \\ ds^2 = -(a\tilde \xi)^2d\tau^2+d\tilde\xi^2+dy^2+dz^2 $$

so the new formulas become: $$ \tilde \rho = \frac{1}{a}e^{a\rho}\\ ds^2=-(a\tilde\rho)^2d\tau^2+d\rho^2+\tilde \rho^2\cosh^2(a\tau)d\theta^2+\tilde\rho^2\cosh^2(a\tau)\sin^2\theta d\phi^2 $$

Hope this helps.